Question: Let $f$ be a transformation from $\mathbb{R}^2$ to $\mathbb{R}^2$. Its Jacobian matrix is given below. $J(f) = \begin{bmatrix} \cos(\theta) & -r\sin(\theta) \\ \\ \sin(\theta) & r\cos(\theta) \end{bmatrix}$ Find the Jacobian determinant of $f$. $|J(f)| = $ How will $f$ expand or contract space around the point $r = \dfrac{1}{2}, \theta = \dfrac{\pi}{2}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Leave it the same (Choice B) B Expand it finitely (Choice C) C Contract it finitely (Choice D) D Contract it infinitely
Answer: The Jacobian determinant is the determinant of the Jacobian matrix. It represents the factor by which the transformation $f$ expands or contracts volume around a certain input. $\begin{aligned} |J(f)| &= \det \left( \begin{bmatrix} \cos(\theta) & -r\sin(\theta) \\ \\ \sin(\theta) & r\cos(\theta) \end{bmatrix} \right) \\ \\ &= r \cos^2(\theta) + r \sin^2(\theta) \\ \\ &= r (\cos^2(\theta) + \sin^2(\theta)) \\ \\ &= r \end{aligned}$ If we evaluate $|J(f)|$ at $r = \dfrac{1}{2}, \theta = \dfrac{\pi}{2}$, we get $\dfrac{1}{2}$. Because the Jacobian determinant here has an absolute value less than $1$ but not $0$, we can conclude that $f$ will finitely contract the space around $r = \dfrac{1}{2}, \theta = \dfrac{\pi}{2}$. To recap, the Jacobian determinant of $f$ is $r$, and $f$ will finitely contract the space around the point $r = \dfrac{1}{2}, \theta = \dfrac{\pi}{2}$.